This post contains the answers to this week's Sunday Afternoon Maths and some extension problems based around the originals.

#### Arccos + Arcsin

Let

This means that

This gives us this triangle, as cosine is adjacent ÷ hypotenuse.

*y*= arccos(*x*)This means that

*x*= cos(*y*)This gives us this triangle, as cosine is adjacent ÷ hypotenuse.

Call the angle at the top of the triangle

Sine is opposite ÷ hypotenuse, so sin(

So

Angles in triangle add to

Hence, arcsin(

Therefore arccos(

*z*Sine is opposite ÷ hypotenuse, so sin(

*z*) =*x*So

*z*= arcsin(*x*)Angles in triangle add to

*π*, so*z*=^{π}/_{2}-*y*Hence, arcsin(

*x*) =^{π}/_{2}-*y*Therefore arccos(

*x*) + arcsin(*x*) =*y*+^{π}/_{2}-*y*=^{π}/_{2}##### Extension

What are the values of arcsec(

*x*) + arccosec(*x*) and arctan(*x*) + arccot(*x*)?*i*^{ i}

By Euler's formula,

This means that

=

=

*i*=*e*^{i π/2}.This means that

*i*^{ i}= (*e*^{i π/2})^{i}.=

*e*^{i 2 π/2}.=

**.***e*^{-π/2}It is notable that this is a real number.

##### Extension

There are other values of

*i*to the power of*i*. What are they?*x*^{xxxxx...}

2 =

So x =

*x*^{xxxxx...}= x^{2}.So x =

**√2**.##### Extension

If

*x*^{xxxxx...}is equal to four, what is the value of*x*?#### 4 and 5 digit numbers

Adam must have removed the final digit of his number before adding or the final digit of the sum would have been an even number. If his original number was ABCDE then, using this, we have:

52713 = ABCDE + ABCD = 11 × ABCD + E.

52713 ÷ 11 = 4792 remainder 1 so Adam's original number was 47921 which has a digit sum of

**23**.
I don't think there are any other values of i^i.

ReplyDeleteI was assuming that you had the R x [0,2pi] definition of (r, theta) to define C. I guess you can have e^(pi/2 + k*2*pi) solutions if you define it on the non-bijective R x R definition of (r, theta).

ReplyDeleteI don't like your triangle proof - what happens when y is outside (0, Π\2) you can't have negative sides to a triangle!

ReplyDelete